Descripción: PDT 621 - NORMAS, CRACTERISTICAS, CASO PRACTICOS
FORMATO DEL PDT 621. EN EXCELDescripción completa
Caso Practico PDT 621 IGV_RentaDescripción completa
20. Very pure liquid water can be subcooled at atmospheric pressure to temperatures well below 273.15 K (0°C). Assume that 1 kg has been cooled as liquid to 267.15 K (-6°C). A small ice crystal (of negligible mass) is added to “seed” the subcooled liquid. If the subsequent change occurs adiabatically at atmospheric pressure, what fraction of the system freezes and what is the final temperature? What is ∆Stotal for the process, and what is its irreversible feature? The latent heat of fusion of water at 273.15 K (0°C) is 333.4 J g -1, and the specific heat of subcooled liquid water is 4.226 J g-1 °C-1. Penyelesaian : Diketahui : T1 = -6oC m = 1 kg ∆H = 333.4 J/g Cp = 4.226 J/goC Proses Adibatik pada Tekanan Konstan Isenthalpic ∆Htotal = 0 Maka : a. Fraksi massa Cp. ∆T + x. ∆H = 0
20. The state of 1 kg of steam is changerd from saturated vapor at 1.38 bar to superheated vapor at 15 bar and 811.15 K (538°C). What are the enthalpy and entropy changes of the steam? What would the enthalpy and entropy changes be if steam were an ideal gas? Penyelesaian : Diketahui : P1 = 20 psia
P2 = 50 psia T2 = 1000oF
Saturated Vapor
Superheated Vapor
Ditanyakan : ∆H & ∆S Mencari nilai entalpi dan entropi masing-masing kondisi dari Tabel F.3 & F.4 : T1 = 228 oF
S = 1.732 H1 = 1156 1
S = 2
H2 = 1534.3 2
∆H = H2-H1 = 1534.3-1156 = 378.3
∆S = S2-S1 = 2 - 1.732 = 0.268
Jika menggunakan persamaan gas ideal : T1 = 385.55 K T2 = 814.44 K ∆H =
